Ten different books are arranged on a shelf . Find the number of different ways in which this can be done. If two specified book are (a) to be together, (b) not to be together.

Answer

To solve the problem of arranging 10 different books on a shelf with specific conditions regarding two specified books, we will break it down into two parts: (a) when the two specified books are to be together, and (b) when the two specified books are not to be together.

### Part (a): Two Specified Books to be Together

1. **Consider the Two Books as One Unit**:
Since the two specified books need to be together, we can treat them as a single unit or block. This means instead of arranging 10 individual books, we now have 9 units to arrange (the block of 2 books + the other 8 books).

2. **Calculate the Arrangements of the Units**:
The number of ways to arrange these 9 units is given by \(9!\) (factorial of 9).

3. **Arrange the Two Books within the Block**:
The two specified books can be arranged among themselves in \(2!\) (factorial of 2) ways.

4. **Total Arrangements**:
Therefore, the total number of arrangements when the two specified books are together is:
\[ \text{Total Arrangements} = 9! \times 2! \]

### Part (b): Two Specified Books Not to be Together

1. **Calculate Total Arrangements Without Restrictions**:
First, we calculate the total arrangements of the 10 books without any restrictions, which is \(10!\).

2. **Calculate Arrangements When the Two Books are Together**:
As calculated in part (a), the number of arrangements when the two specified books are together is \(9! \times 2!\).

3. **Use Complementary Counting**:
To find the number of arrangements where the two specified books are not together, we subtract the arrangements where they are together from the total arrangements:
\[ \text{Arrangements Not Together} = 10! - (9! \times 2!) \]

4. **Simplify the Expression**:
We can factor out \(9!\) from the expression:
\[ 10! = 10 \times 9! \]
Thus,
\[ \text{Arrangements Not Together} = 10 \times 9! - 2 \times 9! = (10 - 2) \times 9! = 8 \times 9! \]

### Final Answers:
- (a) The number of arrangements when the two specified books are together is \(9! \times 2!\).
- (b) The number of arrangements when the two specified books are not together is \(8 \times 9!\).